a, A xác định khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\x-\sqrt{x}\ne0\\3\sqrt{x}+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
b, \(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{1+\sqrt{x}}+\frac{2}{x-1}\right)\)
\(=\left[\frac{\left(\sqrt{x}\right)^2}{\left(\sqrt{x}-1\right)\sqrt{x}}-\frac{1}{\left(\sqrt{x}-1\right)\sqrt{x}}\right]:\left[\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\frac{x-1}{\left(\sqrt{x}-1\right)\sqrt{x}}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1}{\sqrt{x}-1}=\frac{x-1}{\sqrt{x}}\)
c, \(A>0\Leftrightarrow\frac{x-1}{\sqrt{x}}>0\Leftrightarrow x-1>0\Leftrightarrow x>1\)
